Problem: $\overline{AC}$ is $4$ units long $\overline{BC}$ is $6$ units long $\overline{AB}$ is $2\sqrt{13}$ units long What is $\sin(\angle BAC)$ ? $A$ $C$ $B$ $4$ $6$ $2\sqrt{13}$
Solution: SOH CAH TOA in = pposite over ypotenuse opposite $= \overline{BC} = 6$ hypotenuse $= \overline{AB} = 2\sqrt{13}$ $\sin(\angle BAC)=\frac{6}{2\sqrt{13}}$ $=\dfrac{3\sqrt{13} }{13}$